A．logx＝(－k/2.303)t＋logx₀
B．logC＝(－k/2.303)t＋logC₀
C、logC′＝(－k/2.303)t′＋log(k₀/vk)
D．logC′＝(－k/2.303)t′＋log[k₀(1－E^－kt)/vk]
E．C＝k₀(1－E^－kt)/kv 单室模型静脉滴注给药过程中体内血药浓度随时间变化的关系式

根据网考网移动考试中心的统计，该试题：

A．logx＝(－k/2.303)t＋logx0B．logC＝(－k/2.303)t＋logC0C、logC′＝(－k/2.303)t′＋log(k0/vk)D．logC′＝(－k/2.303)t′＋log[k0(1－E－kt)/vk]E．C＝k0(1－E－kt)/kv 单室模型静脉滴注给药过程中体内血药浓度随时间变化的关系式